Lecture 11
The Hydrogen atom
97
98
11.1
LECTURE 11. THE HYDROGEN ATOM
Introduction
The H atom is a bound state of a proton and an electron. The masses of the two particles
are respectively:
mp = 1.7 × 10−27 kg ,
−30
me = 0.91 × 10
kg .
(11.1)
(11.2)
They have opposite charges, q and −q, with:
q = 1.6 × 10−19 C .
(11.3)
mp /me = 1836.15267247(80) ;
(11.4)
The ratio of the two masses is:
it is known to 5 parts per billion!!
The interaction between the two particles is due to electromagnetism; in a nonrelativistic
formulation we can therefore model the H atom as a particle of reduced mass m:
1
1
1
1
=
+
≈
,
µ
mp me
0.995me
(11.5)
in a Coulomb potential:
V (r) = −
q2
e2
≡− .
4π�0 r
r
(11.6)
Given that the proton mass is much larger than the electron one, the reduced mass of the ep
system is very close to the electron mass. The distance r that appears in the expression for
the Coulomb potential is the distance between the electron and the proton. We can identify
the origin of our reference frame with the position of the proton; the potential is clearly
symmetric under rotations around the origin. The parameters that specify the physical
system are the reduced mass µ and the electron charge e.
The H atom is an example of motion in a central force, and we are going to use the
formalism developed in Lecture 10 to find the stationary states of this system.
11.2
Stationary states
The Hamiltonian for the system is:
Ĥ = −
�2 2 e 2
∇ −
.
2µ
r
(11.7)
11.2. STATIONARY STATES
99
The time-independent Schrödinger equation
Ĥψ(r) = Eψ(r) ,
(11.8)
is solved as usual by separation of variables. Using for the solution ψ(r) the Ansatz:
ψ(r, θ, φ) =
χnl (r) m
Y� (θ, φ) ,
r
we obtain the radial one-dimensional equation:
� 2 2
�
� d
�2 �(� + 1) e2
−
+
−
χnl (r) = Enl χnl (r) .
2µ dr2
2µr2
r
(11.9)
(11.10)
Eq. (11.10) is simply Eq. (10.7) for the specific case of the Coulomb potential. The effective
potential in this case is:
e2
�2 �(� + 1)
Vl (r) = − +
.
(11.11)
r
2µ r2
The effective potential is sketched in Fig. 11.1. The boundary condition for the radial function
χnl (r) is:
χnl (0) = 0 .
(11.12)
Remember that, since we are looking for the bound states of the system, we are only interested
in solutions with negative energy, i.e. Enl < 0. Eq. (11.10) suggests that the energy levels
will depend on the total angular momentum �.
It is useful to describe the solutions of the Schrödinger equation in terms of two physical
quantities; a characteristic length:
a0 =
�2
≈ 0.52 Å ,
µe2
(11.13)
and energy:
µe4
≈ 13.6 eV .
2�2
the energy levels, and introduce the dimensionless variables:
�
ρ = r/a0 ,
λnl = −Enl /EI .
EI =
Let us denote by Enl
(11.14)
(11.15)
Eq. (11.10) can be written as:
�
−
�
�2 1 d 2
�2 �(� + 1)
e2
+
−
−
E
nl χnl (a0 ρ) = 0 .
2µ a20 dρ2
a0 ρ
2µa20 ρ2
(11.16)
100
LECTURE 11. THE HYDROGEN ATOM
Figure 11.1: Effective potential for the one-dimensional radial Schrödinger equation for a
system with total angular momentum l. We look for solutions of the time-independent
Schrödinger equation with negative energy E corresponding to bound states of the proton/electron system.
Introducing
un� (ρ) = χn� (a0 ρ) ,
Eq. (11.16) can be rewritten as:
� 2
�
d
�(� + 1) 2µe2 a0 1 2µa20
−
+
+ 2 En,� un� (ρ) = 0 ,
dρ2
ρ2
�2 ρ
�
and finally
�
�
d2
�(� + 1) 2
2
−
+ − λnl un� (ρ) = 0 .
dρ2
ρ2
ρ
(11.17)
(11.18)
(11.19)
We need to solve Eq. (11.19) in order to find the eigenvalues of the Hamiltonian Enl and the
corresponding radial wave functions unl (r). Remember that the angular part of the wave
functions is given by the spherical harmonics.
11.3. SOLUTION OF THE RADIAL EQUATION
11.3
101
Solution of the radial equation
In this section we shall discuss some technical details related to the solution of the radial
equation (11.19). We wish to keep the mathematical details of the solution separated from
the physical interpretation, which will be discussed in the next section.
In order to shine a light on the form of the solution, we can start by considering its
limiting behaviours as ρ → 0 and ρ → ∞.
Let us first discuss the large-distance regime. As ρ is increased, both the centrifugal and
Coulomb potentials tend to zero and become unimportant in Eq. (11.19), which becomes:
�
�
d2
2
− λnl unl (ρ) = 0 .
dρ2
(11.20)
The solutions to this latter equation are simply:
unl (ρ) = exp(±λnl ρ) ,
(11.21)
and the solution that grows exponentially must be discarded because it yields a non-normalizable
wave function.
Of course we cannot completely neglect the potential terms, so we will look for a complete
solution of the form:
unl (ρ) = e−λnl ρ ηnl (ρ) .
(11.22)
Eq. (11.19) becomes:
�
�
�(� + 1) 2
��
�
ηnl
− 2λnl ηnl
+ −
+
ηnl = 0 ,
ρ2
ρ
(11.23)
where the prime symbol “�” denotes the differentiation with respect to ρ. The boundary
condition for un� translates into a boundary condition for ηnl , namely ηnl (0) = 0.
As ρ → 0 we know from the discussion in Lecture 10 that unl ∼ ρ�+1 . Therefore we can
look for a solution for ηnl expanded as a power series in ρ:
ηnl (ρ) = ρ�+1
∞
�
cq ρq .
(11.24)
q=0
Inserting this Ansatz into Eq. (11.23) we obtain:
�
q
(q + l + 1)(q + l)cq ρq+�−1 − 2λnl (q + � + 1)cq ρq+� +
+ 2cq ρq+� − �(� + 1)cq ρq+�−1 = 0 ,
(11.25)
102
LECTURE 11. THE HYDROGEN ATOM
and hence:
�
q
q(q + 2� + 1)cq ρq+�−1 − 2 [λnl (q + � + 1) − 1] cq ρq+� = 0 .
(11.26)
Shifting the summation index, q → q − 1, in the second term of the sum above, we obtain:
�
[q(q + 2� + 1)cq − 2 [λnl (q + �) − 1] cq−1 ] ρq+�−1 = 0 .
(11.27)
q
Since the last equality must hold for all values of ρ, we deduce:
q(q + 2� + 1)cq − 2 [λnl (q + l) − 1] cq−1 = 0 .
(11.28)
Eq. (11.28) is a recursion relation between the coefficients of the Taylor expansion of ηnl /ρl+1 .
It is crucial to note that for large q:
cq
cq−1
q→∞
∼
2λnl
,
q
i.e. cq ∼
(2λnl )q
.
q!
(11.29)
The asymptotic behaviour for cq would yield a solution:
ηnl (ρ) ∼ ρ�+1 e2λnl ρ ,
(11.30)
which in turn yields a wave function unl that is not normalizable.
Therefore we must have cq = 0 for some finite value of q, that we will denote q = nr > 0.
According to Eq. (11.28) this can only happen if:
λnl =
1
1
≡ .
nr + �
n
(11.31)
Then the expansion in Eq. (11.24) only contains a finite number of terms, i.e. it is simply
a polynomial in ρ of finite order nr .
We can see from Eq. (11.31) that this condition implies that the energy eigenvalues
are quantized. Remember that λnl is related to the eigenvalues of the Hamiltonian via
Eq. (11.15). Energy quantization is a consequence of having required the wave function to
be normalizable.
11.4
Physical interpretation
The computation in the previous section shows that the eigenvalues of the Hamiltonian for
the H atom are:
−EI
Enl =
,
(11.32)
(nr + �)2
11.4. PHYSICAL INTERPRETATION
103
where � is the angular momentum of the state, and nr > 0 is an integer. We see that for
the H atom the value of the energy does not depend on nr and �, but only on their sum
n = nr + �. The integer n is called the principal quantum number; its value characterizes the
so-called electron shells.
We can rewrite Eq. (11.14) as:
1
EI = α2 µc2 ,
(11.33)
2
where α is the fine-structure constant:
α=
e2
1
�
.
�c
137
(11.34)
Note that µ � me , and therefore µc2 is the rest energy of the electron. Eq. (11.33) shows that
the typical scale of the energy levels in the H atom is 10−4 the rest energy of the electron.
This justifies the nonrelativistic treatment of the H atom that we have used here. Clearly
there are relativistic corrections; however they are small effect, typically of order α, and can
be studied in perturbation theory.
Using Eq. (11.13) we can express the energy levels as:
En = −
e2
,
2n2 a0
n = nr + l .
(11.35)
This the formula put forward by Bohr in 1913, before Quantum Mechanics was fully developed.
Eq. (11.35) shows clearly that for a given principal quantum number n, � can take the
values � = 0, 1, . . . , n − 1 (corresponding respectively to nr = n, n − 1, . . . , 1). Since for each
value of � we have a (2� + 1) degeneracy, the total degeneracy of the level En is:
n−1
�
�=0
(2� + 1) = n(n − 1) + n = n2 .
(11.36)
The polynomials wnl are called the associated Laguerre polynomials, their explicit expression can be found in standard textbooks on special functions 1 . The full solution for the
eigenfunctions of the energy is:
ψnlm (r) = Rnl (r)Y�m (θ, φ) .
1
see e.g. M. Abramowitz and I.A. Steglun, Handbook of mathematical functions.
(11.37)
104
LECTURE 11. THE HYDROGEN ATOM
The first few radial functions are:
−3/2 −r/a0
R1,0 = 2a0
e
�
,
(11.38)
�
1 −3/2
r
R2,0 = √ a0
2−
e−r/(2a0 ) ,
a
2 2
0
1 −3/2 r −r/(2a0 )
R2,1 = √ a0
e
.
a0
2 6
(11.39)
(11.40)
It is worthwhile to conclude this Lecture with a few remarks.
• The formula for the energy levels reproduces the Bohr spectrum, in agreement with the
experimental data. The degeneracy of each level can only be obtained by the proper
quantum-mechanical description that we have presented.
• The Bohr radius a0 is the typical spatial extension of the ground state.
• The treatment described here can be applied to any hydrogen-like atom, i.e. an atom
with an electron and a nucleus of charge Zq. Simply replace everywhere e2 → Ze2 .
• It is interesting to compute the expectation value of the momentum. Since the�typical
size of the atom is a0 , we deduce from the Heisenberg uncertainty relation that �p2 � ∼
µe2 /�. Thus we obtain:
v ∼ p/µ � e2 /� = αc �
1
c � c.
137
(11.41)
We see a posteriori that motion of the electron is nonrelativistic, and hence our nonrelativistic description is accurate. Relativistic corrections are expected to be O(v/c) �
O(α).
• The ionization energy is the energy needed to extract the electron from the bound state
and is EI . This is know as the Rybderg energy.
• The electron in the H atom can go from one shell to a lower one by emitting a photon.
The series of transitions from principal number n ≥ 2 to n = 1 is called the Lyman
series 2 . The transitions are names by Greek letters: the transition from n = 2 to
n = 1 is called Lyman-α, from 3 to 1 Lyman-β, etc. Likewise the transitions from
n ≥ 3 to n = 2 form the Balmer series. These transitions, which were discovered
before Quantum Mechanics was established, have been observed with great precision
and provide a strong experimental evidence in favour of Quantum Mechanics.
2
A physics department at Harvard Unviversity is named after Lyman.
11.5. SUMMARY
11.5
Summary
As usual, we summarize the main concepts introduced in this lecture.
• Setup of the problem. Hamiltonian for the H atom.
• Separation of variables. One-dimensional equation for the radial wave function.
• Effective potential, boundary condition.
• Solutions for the stationary states, quantization of the energy levels.
• Discussion of the mathematical results. Physical picture.
105
106
LECTURE 11. THE HYDROGEN ATOM